A chain of interacting particles under strain

Stochastic Processes and their Applications 121 (2011) 2014–2042 www.elsevier.com/locate/spa

A chain of interacting particles under strain Michael Allman ∗ , Volker Betz, Martin Hairer Department of Mathematics, University of Warwick, Coventry, CV4 7AL, England, United Kingdom Received 20 July 2010; received in revised form 13 February 2011; accepted 11 May 2011 Available online 20 May 2011

Abstract We investigate the behaviour of a chain of interacting Brownian particles with one end fixed and the other end moving away at slow speed ε > 0, in the limit of small noise. The interaction between particles is through a pairwise potential U with finite range b > 0. We consider both overdamped and underdamped dynamics. c 2011 Elsevier B.V. All rights reserved. ⃝ MSC: 60J70; 60H10; 34E15 Keywords: Interacting Brownian particles; Pitchfork bifurcation; Stochastic differential equations; Singular perturbation

1. Introduction The behaviour of a Brownian particle moving in a potential well and acted upon by a linearly increasing force is widely used to model the mechanical failure of molecular bonds arising in dynamic force spectroscopy experiments [19,9,20,13]. This began with the work of Bell [4] and was developed further by Evans and Ritchie [10]. Let qs denote the length at time s of a bond that is fixed at one end and has a harmonic spring attached to the other. If the spring moves linearly at speed ε > 0, the motion of qs is typically modelled according to an SDE of the form dqs = (−U ′ (qs ) + εs) ds + σ dWs , where U (q) denotes the bond energy (e.g. Lennard-Jones potential), Ws is a standard Brownian motion and σ > 0 is the (small) noise intensity. Note that this model assumes the motion is ∗ Corresponding author.

E-mail addresses: [email protected] (M. Allman), [email protected] (V. Betz), [email protected] (M. Hairer). c 2011 Elsevier B.V. All rights reserved. 0304-4149/$ - see front matter ⃝ doi:10.1016/j.spa.2011.05.007

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overdamped. Rupture of the bond corresponds to the first time qs escapes from the stable well of the effective, time-dependent potential, H (q, εs) = U (q) − εs q. The effect of the external force is to lower the barrier height of H , thus making escape more likely. The main objective is to study the distribution of first-breaking times and how the mean first-breaking time scales with the pulling speed ε. Typically, two pulling speed regimes are considered. For very slow pulling, the particle is able to escape the well through a large deviation event before the potential has changed significantly and the energy barrier is still large. In order to apply the standard theory valid for time-independent potentials [16,11,7,12], the adiabatic approximation is used: at any given time s, the bond has an instantaneous rate of rupture, ˙ k(s), and the probability of survival until time s, denoted P(s), decays according to P(s) = −k(s)P(s). Note that 1/k(s) is given by the usual Eyring–Kramers formula [16,11,7] applied to H at time s. As the speed of pulling increases, the energy barrier at the time of rupture becomes smaller. If pulling is sufficiently fast, the barrier may be close to vanishing completely when rupture occurs. This means that the external force, given by εs, is almost equal to the maximum slope of U , which occurs at the point of inflection between its minimum and maximum, i.e. maximum slope is U ′ (c0 ), where U ′′ (c0 ) = 0. For times s at which εs is close to this critical force, the effective potential H is almost cubic near its minimum. This leads to a different rupture rate than that given above, although still calculated within the Kramers framework. It is interesting to consider what happens as the pulling speed increases yet further and the Eyring–Kramers formula is no longer applicable, nor the adiabatic approximation underpinning the above approach. In this paper, we consider this situation in a model related to that above. More precisely, we consider a chain of two identical bonds in series with one end fixed and the other being pulled at a constant rate ε. Both overdamped and underdamped dynamics are treated. We are interested in which of the two bonds breaks first and how this depends on ε and the noise intensity σ . As above, the dynamics near the inflection point of the bond energy U play an important role and will be the focus of our analysis. Roughly, we find that for ε > σ 4/3 , the right-hand bond breaks first, while for ε < σ 4/3 , both have an equal probability of breaking in the limit of small noise. Thus ε = σ 4/3 represents the threshold at which the adiabatic approximation becomes valid. To our best knowledge, the first work to tackle rigorously such models of bonds under an external, time-dependent force was [2]. There the authors consider a similar model of two bonds in series as above, but with an additional assumption that U is cut-off strictly convex. The breaking event corresponds to the first time one of the two bonds exceeds the range of U . Roughly speaking, it is shown that for ε > σ , the chain always breaks on the right-hand side, whereas for ε < σ , each bond has an equal chance to break in the small noise limit. Thus the threshold between the different types of behaviour is different from that found in the present work, where the bond energy U is taken to be smooth (but also with finite range). In principle, the results of [2] can be extended to arbitrarily many bonds in series [1]. The behaviour of several bonds in series has also been considered by many authors, for both time-dependent and time-independent external forces. The situation when the external force is constant, i.e. one initially stretches the chain by some amount and then fixes both endpoints, has been considered for harmonic potentials [18] and Lennard-Jones potentials [21]. In the harmonic case, it is shown analytically and numerically that the probability to break at either endpoint is half that of breaking at any non-extremal point, which all have the same probability. In the Lennard-Jones case, the motion is not assumed to be overdamped, i.e. the authors consider the equation

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m q¨i (s) = −mγ q˙i (s) −

 ∂H (q(s)) + 2γ W˙ i (s), ∂qi

2 ⩽ i ⩽ N − 1,

N where the Wi are independent Brownian motions, q(s) = (qi (s))i=1 ∈ R N is the chain configuration, with q1 and q N constant, and H is the total potential energy of the chain. It is assumed beforehand that one bond is close to breaking and all others are close to minimal energy so that a quadratic approximation can be used for H . Then the breakage rate is calculated, alongside simulations, using a multi-dimensional version of Kramers’ theory developed by Langer [17]. These show that the breakage rate is lower the closer the chosen weak bond is to the chain endpoints. However, the simulations also show that the harmonic approximation for H may fail for bonds near the breaking bond, which the authors there suggest may explain some discrepancies between the theory and simulations. As they point out, Langer’s theory, as well as the classical Kramers theory for a single particle, requires a harmonic approximation for H . The case of a chain with one end fixed and a linearly increasing force applied at the other end has been considered by Fugmann and Sokolov in [14,15] to model the mechanical failure N ∈ R N +1 , with of a polymer chain. More precisely, they consider the vector q(s) = (qi (s))i=0 q0 ≡ 0, which evolves according to the SDE

dqi (s) = −

∂H (q(s), εs) ds + σ dWi (s), ∂qi

1 ⩽ i ⩽ N,

where H (q, εs) is the time-dependent potential energy of the chain, given by − H (q, εs) = U (qi − q j ) − εs q N , 0⩽i< j⩽N

and U is the Morse potential. A break is said to occur when q(s) overcomes an energy barrier of the effective potential, H . Numerically, they show that for a high pulling speed, ε, only the right half of the chain contributes to the breaking event and the probability increases as you move towards the right. For smaller ε, they show that the breakpoint is more uniformly distributed along the whole chain. In their analysis, they assume that the rupture dynamics of different bonds are independent and then apply the one-dimensional theory. The only thing left to do then is to analyse how much force each bond feels, which depends on the pulling speed. For lower pulling speeds, it is assumed that each bond feels the same force, whereas for higher pulling speeds the chain is approximated by a harmonic chain. This paper is organised as follows. In Section 2, we introduce our model and deduce Eq. (2.2), which is our main object of study. Our results are then stated in Theorems 2.1–2.3. In Sections 3–5 we give the proofs. The following notation is used in this paper: • By x1 (t, ε) ≍ x2 (t, ε) we mean that for two functions x1 (t, ε), x2 (t, ε), defined for t in an interval I and 0 < ε ⩽ ε0 , there exist constants c± > 0 such that c− x2 (t, ε) ⩽ x1 (t, ε) ⩽ c+ x2 (t, ε) for all t ∈ I and 0 < ε ⩽ ε0 . • By ox (1) we mean that limx→0 ox (1) = 0. • x1 . x2 means that there exists a constant c > 0 such that cx1 ⩽ x2 for x1 , x2 > 0 sufficiently small. • x1 ≪ x2 means that x1 = o(x2 ). • We shall write Pt0 ,q0 to denote probability conditioned on the relevant process starting at time t0 in position q0 .

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2. The model and main results Three particles, q L , q and q R in R, interact with each other via a pairwise potential U . We assume that U is smooth with finite range b > 0 and a unique minimum at 0 < a < b, with U ′′ (a) > 0. We also assume that there is a unique c0 ∈ (a, b) such that U ′′ (c0 ) = 0. The particle q L is fixed at the origin and the position of q R at time s ⩾ 0 is given by q R (s) = 2a(1 + εs), where ε > 0 is a small parameter. We study the behaviour of the middle particle, with position at time s given by qs . Initially, it has position q0 = a so that the distance between neighbouring particles is a. The middle particle evolves according to an SDE of the form dqs = ps ds, ε β−1 d ps = − ps ds −

∂H ( ps , qs , εs) ds + σ dWs , ∂q

where Ws is a standard one-dimensional Brownian motion with W0 = 0, σ > 0 is the noise intensity, β ∈ R and H ( p, q, εs) is given by p2 + U (q) + U (2a(1 + εs) − q). 2 Rescaling time as t = εs, this is the same in law as solving H ( p, q, εs) =

1 pt dt, ε 1 1 ∂H σ ε β−1 d pt = − pt dt − ( pt , qt , t) dt + √ dWt . ε ε ∂q ε

dqt =

(2.1)

The length of the chain (q L , q, q R ) increases linearly with t. Clearly, if we wait a long enough time, the distance between q and at least one of its neighbours must become greater than the range of U . In this case, these particles no longer interact and the chain can be considered broken. Since U has a minimum at a, it is energetically preferable for q to move towards either q L or q R . Letting ε = ε(σ ), our aim is to determine how the speed of pulling affects which of these two possibilities occurs in the limit as σ ↓ 0. We easily check that the configuration of equally spaced particles satisfies ∂q H = 0, ∂q2 H > 0 until time t0 , where a(1 + t0 ) = c0 . Thus until this time it is a stable configuration and so we expect qt0 ≈ a(1 + t0 ). For t > t0 , this configuration becomes unstable and new minima emerge. So we expect qt to quickly move away from the chain midpoint and towards one of these newly formed minima. Note that as a function of q, H is symmetric about q = a(1 + t), but its time-dependence introduces asymmetry as we shall see. Once qt has approached one of these new minima, we expect it to stay there as the energy barrier to escape becomes higher. The evolution of the chain, therefore, is determined by its behaviour around the bifurcation of H at t = t0 , which we shall now consider. Letting z t = a(1 + t) − qt , we express the term ∂ H/∂q appearing in (2.1) in terms of z. By a Taylor expansion in space, we find ∂H ( pt , qt , t) = U ′ (qt ) − U ′ (2a(1 + t) − qt ) ∂q = U ′ (a(1 + t) − z t ) − U ′ (a(1 + t) + z t ) 1 ≈ −2U ′′ (a(1 + t))z t − U (4) (a(1 + t))z t3 . 3

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Assuming that there is 0 < T < t0 such that for t ∈ [t0 − T, t0 + T ], U (4) (a(1 + t)) is negative and bounded away from zero (see comment below), we have by a Taylor expansion in time, 1 −2U ′′ (a(1 + t))z t − U (4) (a(1 + t))z t3 ≈ 2a(t − t0 )z t − C z t3 . 3 We remark that this assumption about U (4) (a(1 + t)) should not have much effect. At most, the right-hand side above would have a “+C z t3 ” term appearing, but in either case this term is very small for z and t − t0 close to zero, which is where most of our analysis will take place. Making the space and time transformations q = a(1 + t) − q, p = a − p/ε and t = t − t0 , as well as normalising constants to one, we arrive at the SDE dqt = pt dt, 1 σ ε β d pt = − pt dt + (tqt − qt3 + ε) dt + √ dWt . ε ε

(2.2)

This will be our main equation for the rest of this paper. By the above discussion, understanding how its solution behaves will be a good indication of the behaviour of the original chain. Eq. (2.2) represents the motion of the particle q in the potential (1/ε)V (q, t) := (1/ε)(− 21 tq 2 + 14 q 4 ) with an additional +1 force giving the particle a small bias towards the right. This force comes from pulling the chain (q L , q, q R ) and corresponds to the fact that in the absence of noise, q does not just stay at the chain midpoint a(1 + t), but lags behind by a small amount. Rephrasing the discussion after (2.1), the function V represents the energy of a given chain configuration. For negative times, the origin, corresponding to equally spaced particles, minimises V . When t = 0, V undergoes a symmetric pitchfork bifurcation at the origin. For √ √ positive times, V has two minima located at ± t. For t > 0 large enough these minima at ± t correspond to the configurations where q is a distance a from q L or q R , respectively, and more than b from the other. In terms√of (2.2),√the aim of this paper can be roughly stated as to determine whether q moves towards + t or − t as t becomes positive, which corresponds to the chain ‘breaking’, and how it is affected by the speed of pulling. There are several ways in which one may rigorously define the chain to break. One possible definition is that the chain breaks as soon as the distance between q and one of its neighbours exceeds the range of the pairwise potential U . Then the chain either breaks on the right- or lefthand side, depending on whether q R − q > b or q − q L > b, respectively. This was used in [2]. Alternatively, one may consider the chain to break as soon as the chain configuration reaches a neighbourhood of one of the energy minima that emerges after the √bifurcation. In the above formalism, this means the process qt reaching a neighbourhood of ± t. We shall avoid making this choice by instead giving a precise description of the behaviour of qt that contains more information than any of these possible definitions. Eq. (2.2) in full is not something we can treat. But there are two obvious simplifications: the first is to omit the qt3 term in the equation for p, leading to a linear equation that can be solved explicitly; the second is to neglect mass, taking ε β d p = 0, and consider the overdamped equation. We treat both of these and obtain satisfactory results. Taking σ = ε α+1/2 for α > −1/2, we firstly consider the linear SDE dqt0 = pt0 dt, 1 ε β d pt0 = − pt0 dt + (tqt0 + ε) dt + ε α dWt . ε

(2.3)

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Denoting by Ps the law of the solution with vanishing initial condition at time s < 0, we have the following result. Theorem 2.1. Let qt0 be the solution of (2.3). If α > 1/4 then   lim lim inf Ps lim qt0 = +∞ = 1, ε↓0 s→−∞

t→∞

while if α < 1/4 then     lim lim inf Ps lim qt0 = +∞ = lim lim inf Ps lim qt0 = −∞ = 1/2. ε↓0 s→−∞

t→∞

ε↓0 s→−∞

t→∞

This theorem shows that the threshold between fast and slow pulling regimes is given by α = 1/4 and is independent of β. However, we also note that if we start the processes at a finite negative time −T , then neglecting mass does have an effect, but only for β < 0. Indeed, for −1 < β < 0 and zero initial conditions, the threshold becomes α = (1 + β)/4 (see [1]). This result has some clear limitations: for t > 0, qt0 shoots off quickly to ±∞ as the drift becomes ever more repelling, while the solution of (2.2) is prevented from doing this by the nonlinear term and so is more likely to return to the origin. Secondly, we neglect the mass term εβ d p in (2.2) and consider the one-dimensional overdamped equation dqt =

1 σ (tqt − qt3 + ε) dt + √ dWt . ε ε

(2.4)

Again letting Ps denote the law of the solution with vanishing initial condition at time s < 0, we have √ Theorem 2.2. Let qt solve (2.4). There exist constants c1 , γ > 0 such that if t1 = c1 ε| ln σ | then (1) (Fast Pulling) for any σ 4/3 | ln σ |2/3 ≪ ε(σ ) ≪ 1,   qt lim lim inf Ps inf √ > γ = 1. t1 ⩽t σ ↓0 s→−∞ t (2) (Slow Pulling) for any σ 2 | ln σ |3 . ε(σ ) ≪ σ 4/3 | ln σ |−13/6 ,      s 1  qt  lim lim sup P inf √ > γ −  = 0 t1 ⩽t σ ↓0 s→−∞  2 t and     qt  s lim lim sup P sup √ < −γ − σ ↓0 s→−∞  t t1 ⩽t

 1   = 0. 2

As will be clear from the proof of this theorem, the result also holds if we replace the ε in brackets in (2.4) by a term of the form kε for any constant k > 0. This will be used in the proof of Theorem 2.3 below. Letting σ = ε α+1/2 above gives α = 1/4 as the threshold between the different regimes, as found in Theorem 2.1. We also remark that the threshold between fast and slow pulling regimes here differs from that in [2], where it is roughly ε = σ . This difference can be attributed to the bifurcation of V : for t near zero, V is almost flat and so the additional +1 force requires slower

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pulling, or stronger noise, to be counteracted than it does in [2], where the potential has positive curvature bounded away from zero. Having considered two simplifications of (2.2), we finally return to the full solution itself. Intuitively, by taking β large, the effect of the mass term ε β should become small and the solution should behave like that of the overdamped equation (2.4). So if we show that for suitably large β, the difference between the two solution stays small, we can use Theorem 2.2 to tell us about (2.2). This leads us to the following result. Theorem 2.3. Let qt solve (2.2) √ with β > 2. There exist constants c1 , γ > 0, independent of β and σ , such that for t1 = c1 ε| ln σ | and any t2 > t1 , (1) (Fast Pulling) if σ 4/3 | ln σ |2/3 ≪ ε(σ ) ≪ 1 then   qt s lim lim inf P inf √ > γ = 1, t1 ⩽t⩽t2 σ ↓0 s→−∞ t (2) (Slow Pulling) while if 0 < δ < β/2 − 1 and σ 2/(1+2δ) ≪ ε(σ ) ≪ σ 4/3 | ln σ |−13/6 then      s  qt  lim lim sup P inf √ > γ − 1/2 = 0 t1 ⩽t⩽t2 σ ↓0 s→−∞ t and       qt  s  sup √ < −γ − 1/2 = 0. lim lim sup P  σ ↓0 s→−∞  t t1 ⩽t⩽t2 Note that we only consider finite time intervals here and that there is a slight difference between the lower bound on ε in (2) above and in Theorem 2.2(2). This second point is related to the fact that the mass is of the form ε β . However, it does not affect the threshold between the two regimes. The idea for the proof of Theorem 2.3 is to first show that the solution qt of (2.2) is bounded above and below by two solutions qt+ and qt− of the overdamped equation (2.4), but with the offset ε in the drift term replaced by ε(1 ± r (σ )), where r (σ ) is a term satisfying r (σ ) ≪ 1. Above, qt+ and qt− use the same driving Brownian motion as qt , and by comparing their drift terms, it follows that in the slow pulling case of the above theorem, sample paths of qt+ and qt− must almost surely be attracted to the same potential well as σ ↓ 0. This determines the behaviour of qt . We finally note that (2.2) is not suitable for considering the chain ‘breaking’ due to a large deviation event. In that case, our expansion of the potential is not valid. 3. The linear model In this section, we give the proof of Theorem 2.1 and are therefore dealing with Eq. (2.3). This equation is simple enough to have an explicit solution in terms of Airy functions Ai(t), Bi(t) (see [8]): using the fact that both Ai(z) and Bi(z) solve the equation w ′′ (z) = zw(z), and that the Wronskian Ai(t)Bi′ (t) − Bi(t)Ai′ (t) = 1/π, it can be checked that the process  ∫ t 1 −β qt0 = π ε (1−2β)/3 −Ai(t (ε, β)) e− 2 (t−s)ε Bi(s(ε, β))(ds + ε α dWs ) −∞

+ Bi(t (ε, β))

∫

t

e −∞

− 12 (t−s)ε −β

 Ai(s(ε, β))(ds + ε dWs ) α

(3.1)

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is the (almost surely unique) solution of (2.3) with zero initial conditions and starting time sent to −∞. Above, we have set s(ε, β) = ε −(1+β)/3 (s + ε 1−β /4). By using the asymptotic expansions 1 3/2 Ai(s) = √ s −1/4 e−2s /3 (1 + O(s −2/3 )), 2 π 1 3/2 Bi(s) = √ s −1/4 e2s /3 (1 + O(s −2/3 )), π

s>1

(3.2)

s>1

(3.3)

1 Ai(s) = √ |s|−1/4 cos(2|s|3/2 /3 − π/4) + O(|s|−7/4 ), π

s < −1

(3.4)

−1 Bi(s) = √ |s|−1/4 sin(2|s|3/2 /3 − π/4) + O(|s|−7/4 ), π

s < −1.

(3.5)

The behaviour of qt0 as t → ∞ can be described in a straightforward way by considering the ‘renormalised process’ 1

−β

e 2 tε q˜t = q 0. π ε (1−2β)/3 Bi(t (ε, β)) t 1

Indeed, we then have the following result. Proposition 3.1. q˜∞ := limt→∞ q˜t exists almost surely and is a Gaussian random variable with mean m = m(ε) = ε (1+β)/3 e− 12 ε 1

1−2β

(3.6)

and variance v = v(ε) = ε 2α+(1+β)/3 e− 4 ε

1 1−2β

∫

∞

esε

(1−2β)/3

Ai(s)2 ds.

(3.7)

−∞

Proof. Let us first investigate the deterministic integrals in (3.1). Since Bi(s) < Bi(t) for s < t and t ⩾ 0, we have that the deterministic integral in the first line of (3.1), after renormalisation, t −β is bounded by Ai(t (ε, β)) −∞ esε /2 ds for large enough t, and thus converges to zero as ∞ 3 t → ∞ due to (3.2). It is known [8] that −∞ e ps Ai(s) ds = e p /3 for all p > 0, and thus the limit of the corresponding (renormalised) integral in the second line of (3.1) is given by ∫ ∞ ∫ ∞ s −β 1 1−2β 1 (1−2β)/3 e 2 ε Ai(s(ε, β)) ds = ε (1+β)/3 e− 8 ε e 2 sε Ai(s) ds −∞

−∞

=ε

1 1−2β (1+β)/3 − 12 ε

e

.

Clearly, this is also the limit of E(q˜t ) as t → ∞. The stochastic integrals in q˜t are given by ∫ t ∫ α α J1 (t) = ε h(t) f 1 (s) dWs , J2 (t) = ε −∞

t

f 2 (s) dWs ,

−∞

with h(t) =

Ai(t (ε, β)) , Bi(t (ε, β))

f 1 (s) = Bi(s(ε, β)) esε

−β /2

,

f 2 (s) = Ai(s(ε, β)) esε

−β /2

.

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t By the time change τ1 (t) = ε 2α −∞ f 12 (s) ds, J1 (t) equals h(t)Bτ1 (t) in distribution, where Bs is a standard Brownian motion. By the law of the iterated logarithm, lim sup t→∞

|J1 (t)| =1 √ h(t) 2τ1 (t) ln ln τ1 (t)

(3.8)

almost surely. Again using Bi(s) < Bi(t) for s < t and t ⩾ 0, we find 1/2 ∫ t  sεβ α , ε ds h(t) τ1 (t) ⩽ Ai(t (ε, β))ε −∞

which by (3.2) converges to zero superexponentially fast. By (3.3) it is easy to see that ln ln τ1 (t) grows only proportionally to ln t, and thus the denominator on the left-hand side of (3.8) converges to zero. It follows that J1 (t) → 0 as t → ∞ almost surely. J2 , on the other hand, is a square-integrable martingale, and thus converges almost surely. Each J2 (t) is Gaussian, and thus so is the limit. It has mean zero and variance ∫ ∞ ∫ ∞ −β 2α 2 2α v=ε f 2 (s) ds = ε esε Ai(s(ε, β))2 ds. −∞

−∞

The same change of variable that was employed to get (3.6) yields (3.7).



Since Bi(t) e−t diverges as t → ∞, Proposition 3.1 means limt→∞ |qt0 | = ∞ almost surely. Whether the divergence is to plus or minus infinity is determined by the sign of q˜∞ = limt→∞ q˜t . q˜∞ is a Gaussian random variable with mean m √= m(ε) > 0 given by (3.6), and variance v = v(ε) given by (3.7). Thus if limε→0 m(ε)/ v(ε) = ∞, then limε→0 P(limt→∞ qt0 = +∞) = 1, as √ the distribution of q˜∞ concentrates on the positive half line. On the other hand, if limε→0 m(ε)/ v(ε) = 0, then limε→0 P(limt→∞ qt0 = +∞) = 1/2, as the distribution of q˜∞ becomes spread out and P(q˜∞ > 0) → 1/2 as ε → 0. We now determine the circumstances under which each of the above cases occurs. Define ∫ ∞ J ( p) = e2 ps Ai2 (s) ds. −∞

We have Lemma 3.2. There exist constants c1 and c2 such that 3 (i) lim p→∞ p 1/2 e−2 p /3 J ( p) = c1 , 3 (ii) lim p→0 p 1/2 e−2 p /3 J ( p) = c2 . Proof. Consider first the case p → ∞. Then, ∫ 1 3 3 e2 ps Ai2 (s) dsp 1/2 e−2 p /3 ⩽ C e−2 p /3+2 p p 1/2 → 0 −∞

as p → ∞. For s > 1 we use (3.2) and find ∫ ∞ ∫ ∞ 1 3/2 e2 ps Ai2 (s) ds = e−4s /3+2 ps s −1/2 (1 + O(s −3/2 )) ds 4π 1 1 ∫ ∞ p 3 3/2 = e− p (4t /3−2t) t −1/2 (1 + O( p −3 t −3/2 )) dt, 4π 1/ p2 ∞  1/4  ∞ where we used the substitution s = p 2 t. Decompose the integral as 1/ p2 = 1/ p2 + 1/4 . The first of these is bounded by C e p

3 /3

for some C > 0 and we can ignore it. For the second,

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we have O( p −3 t −3/2 ) = O( p −3 ) and can take this outside the integral. Then by the Laplace method, ∫ ∞ ∫ ∞ 3 3/2 3 3 2 3 e− p (4t /3−2t) t −1/2 dt = e2 p /3 e− p [(t−1) +O(t−1) ] t −1/2 dt 1/4

1/4 2 p 3 /3

= e

p

−3/2 √

π (1 + O(1/ p)).

1 √

Thus (i) holds with c1 = 4 π . For (ii), we use that ∫ ∞ ∫ ∞ e2 ps Ai2 (s) ds → Ai2 (s) ds = const −1

−1

as p → 0. Using (3.4) and ignoring the O(|s|−7/4 ) term, which does not change the resulting expression below, we then get ∫ −1 ∫ 1 −1 2 ps −1/2 2 ps 2 e Ai (s) ds = e |s| cos2 (2|s|3/2 /3 − π/4) ds π −∞ −∞ ∫ ∞ 1 e−2t t −1/2 cos2 (2 p −3/2 t 3/2 /3 − π/4) dt, = √ π p p where in the last line we used the substitution t = − ps. As p → 0, the integral in the last line above converges to √ ∫ 1 ∞ −1/2 −2t π t e dt = √ , 2 0 2 2 which proves (ii) with c2 =

√ 1 3/2 . π2



When substituting p = ε (1−2β)/3 /2 into the last lemma, we find that as ε becomes small, 2β

v(ε) ≍ ε 2α+(1+β)/3 e− 4 ε ε (1−2β)/6 e 12 ε = ε 2α+ 3 + 6 e− 6 ε . √ Thus, m(ε)/ v(ε) ≍ ε −α+1/4 , which completes the proof of Theorem 2.1. 1 1−2β

1

1−2β

1

1 1−2β

4. The overdamped model In this section, we give the proof of Theorem 2.2. As a first step, we show that given a negative time −T ⩽ −1, the process qt starting at 0 at time −∞ will be in X = [−1, 1] with very high probability at time −T . Proposition 4.1. Let X = [−1, 1], let T ⩾ 1, and let qt solve (2.4). Then, we have lim lim inf Ps (q−T ∈ X ) = 1.

σ,ε→0 s→−∞

Proof. Applying Itˆo’s formula to the function q → q 2 , we obtain   d 2t 2 2 4 σ2 1 σ2 2 E(qt ) = E qt − qt + 2qt + ⩽ − E(qt2 ) + ε + , dt ε ε ε ε ε where we made use of the fact that t ⩽ −1. The claim then follows from Chebyshev’s inequality upon letting ε → 0 and σ → 0. 

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The choice X = [−1, 1] is for convenience only; clearly could have chosen any ε-independent interval containing the origin as an interior point. In the second step, we show that the conclusion of Theorem 2.2 holds uniformly over initial conditions belonging to X at time −T . We denote by P−T,x the law of (2.4) with initial condition q−T = x ∈ X . √ Proposition 4.2. Let qt solve (2.4). There exist constants c1 , γ > 0 such that if t1 = c1 ε| ln σ | then (1) (Fast Pulling) for any σ 4/3 | ln σ |2/3 ≪ ε(σ ) ≪ 1,   qt −T,x lim inf P inf √ > γ = 1. t1 ⩽t σ ↓0 x∈X t (2) (Slow Pulling) for any σ 2 | ln σ |3 . ε(σ ) ≪ σ 4/3 | ln σ |−13/6 ,       −T,x qt  lim sup P inf √ > γ − 1/2 = 0 t1 ⩽t σ ↓0 t x∈X

and       qt   −T,x sup √ < −γ − 1/2 = 0. lim sup P  σ ↓0 x∈X  t t1 ⩽t By the Markov property, Propositions 4.1 and 4.2 immediately imply Theorem 2.2. The proof of Proposition 4.2 will be given in the remainder of this section. It relies heavily on ideas and methods developed by Berglund and Gentz in [5,6]. They consider similar equations to (2.4), but with drift terms (1/ε) f (q, t) such that f (q, t) = − f (−q, t) and f (0, 0) = ∂q f (0, 0) = 0, where a stable equilibrium branch undergoes a pitchfork bifurcation and becomes unstable at t = 0. A simple example of such an f is f (q, t) = tq − q 3 . While it is clear by symmetry that in their case the probability of eventually ending up on either stable equilibrium branch after the bifurcation is 1/2, Berglund and Gentz obtain rather precise information on the behaviour of the paths in the bifurcation region. In particular, they find the time t∗ > 0 when a typical path leaves the vicinity of the now-unstable branch q = 0 and approaches one of the stable branches. The present setup differs from that in [5,6] by the asymmetry of the function f with respect to q = 0. Our strategy will be to derive estimates on the paths of the solution to (2.4) that are analogous to those in [5,6]. We will then use these estimates to show that in the case of fast pulling, the asymmetry is strong enough to drive the process towards only one of the two stable equilibrium branches, while in the slow pulling case, the process behaves much as the one without asymmetry. Unfortunately, it turns out that we cannot directly apply the results of [5,6], and that even many of their technical results need to be adapted in order to apply here. While none of the adaptations are particularly difficult, they are lengthy and rather technical. In order not to overburden the paper, we have often chosen to not spell out all the details, but rather to refer to the proofs in [5,6] and give hints as to what needs to be changed to make them work in our situation. Since our notation follows that in [5,6] rather closely, we trust that the reader will be able to fill in the details. 4.1. Fast pulling We begin by considering the fast pulling regime from Proposition 4.2. In this case, the noise in the system is not strong enough to overcome the asymmetry caused by pulling and the qualitative

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behaviour of qt is the same as that of the deterministic solution qtdet of the ODE q˙tdet =

1 det (tq − (qtdet )3 + ε), ε t

det q−T = x.

(4.1)

√ In particular, we will see that qtdet falls into the right-hand well by a time of order ε| ln σ | after the bifurcation and so too does qt . The strategy is as follows: √ √ (1) Show that qt is of order ε when t = ε. √ (2) Show that (qt , t) then leaves the space–time set K(κ) (see (4.8)), by a time of order ε| ln σ |. (3) Show that qt approaches the right-hand well and stays in a small neighbourhood of it up until any time t2 > 0. (4) Show that by taking t2 large enough, qt stays in a neighbourhood of the right-hand well of order t 1/2−γ for any 0 < γ < 1/2 and all t ⩾ t2 . Step One. We begin by describing how qtdet behaves. Lemma 4.3. Let qtdet be the solution of (4.1). Then we have, uniformly for all initial conditions x ∈ X,  √ ε/|t| for −√T + ε| ln √ ε| ⩽ t ⩽ − ε det √ qt ≍ ε for − ε ⩽ t ⩽ ε and there is a constant C > 0 such that for all x ∈ X , |qtdet | < C for all −T ⩽ t ⩽ −T +ε| ln ε|. Proof. Consider the equation ∗ ∗ tq+ (t) − q+ (t)3 + ε = 0.

(4.2)

∗ , we see that for t < 0 it has By fixing t and differentiating the left-hand side with respect to q+ no turning points and so admits a unique real-valued solution. Furthermore, we can check that ∗ (t) ≍ ε and (q ∗ )′ (t) = q ∗ (t)/(3q ∗ (t)2 − t) ≍ ε for negative t bounded away from zero. q+ + + + ∗ (−T ). Define z = q det − q ∗ (t). As Suppose first that the initial condition satisfies x ⩾ q+ t t + long as z t ⩾ 0, we have z˙ t ⩽ t z t /ε so that ∗ ∗ (t) ⩽ (x − q+ (−T )) e(t 0 ⩽ qtdet − q+

2 −T 2 )/2ε

.

Let t0 = −T + ε| ln ε|. If z t < 0 for some −T < t < t0 , which means that qtdet ≍ ε, then the analysis below for t ⩾ t0 can be applied from that time. Otherwise, the above inequality shows that qtdet ≍ ε. 0 ∗ (−T ) then we define z = q ∗ (t) − q det . As (q ∗ )′ (t) > 0 for negative t bounded If x ⩽ q+ t + t + away from zero, we have z t ⩾ 0 for such t. In this case, there is c1 > 0, independent of x ∈ X , such that 1 ∗ z˙ t ⩽ c1 ε + (t z t + 3q+ (t)z t2 ) ε ∗ (t) (which is satisfied by z(−T ) for for all −T ⩽ t ⩽ t0 . Furthermore, as long as z t ⩽ −t/6q+ ∗ 2 all x ∈ X by taking ε sufficiently small) then 3q+ (t)z t ⩽ −t z t /2 and so

z˙ t ⩽ c1 ε +

1 t zt . 2ε

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This tells us 0⩽

∗ q+ (t) − qtdet

(t 2 −T 2 )/4ε

⩽ (q (−T ) − x) e ∗

+ c1 ε

∫

t

e(t

2 −s 2 )/4ε

ds,

−T

∗ (t) for all −T ⩽ t ⩽ t and we can use the above inequality to which shows that z t ⩽ −t/6q+ 0 det again see that qt0 ≍ ε. We now analyse the behaviour for t ⩾ t0 . As qtdet > 0 and q˙tdet = 1 whenever qtdet = 0, it 0 det follows that qt ⩾ 0 for all t ⩾ t0 . Therefore, for t ⩾ t0 we have

q˙tdet ⩽

1 det (tq + ε) ε t

and so qtdet ⩽ qtdet e(t 0

2 −t 2 )/2ε 0

∫

t

+

e(t

2 −s 2 )/2ε

 ds ⩽

t0

√ c2 √ ε/|t| for t√ 0 ⩽ t ⩽ − √ε c2 ε for − ε ⩽ t ⩽ ε

(4.3)

for some constant c2 > 0. To obtain the lower bound, we use that for t ⩽ √ 0, tq − q 3 ⩾ 2tq 2 det as long as q ⩽ |t|. By taking ε sufficiently small, we have 0 < qt0 ⩽ |t0 | for all initial √ conditions x ∈ X . As long as 0 ⩽ qtdet ⩽ |t|, then q˙tdet ⩾

1 (2tqtdet + ε) ε

and qtdet ⩾ qt0 e(t

2 −t 2 )/ε 0

∫

t

+

e(t

2 −s 2 )/ε

ds.

t0

√ √ By (4.3), we certainly have qtdet ⩽ |t| for t ⩽ − ε and so√the above inequality gives the √ det corresponding lower bound for qt up until this time. For − ε ⩽ t ⩽ √ε, we then have tqtdet − (qtdet )3 ⩾ −Cε for some constant C > 0, so that qtdet remains of order ε in this interval. This completes the proof.  √ We now show that the deviation process yt := qt − qtdet satisfies |y( ε)| < hε −1/4 for some h ≪ ε 3/4 , which will complete Step One. The process yt solves dyt =

1 σ [a(t)yt + b(yt , t)] dt + √ dWt , ε ε

y(−T ) = 0,

(4.4)

where a(t) = t − 3(qtdet )2 and b(yt , t) = −3qtdet yt2 − yt3 . For all pairs (y, t) ∈ B(h) for a choice of h = O(ε1/4 ) (see (4.7) and Lemma 4.4), we have |b(y, t)| ⩽ M y 2 . Solving (4.4) gives ∫ t ∫ σ 1 t α(t,s)/ε α(t,s)/ε yt = √ e dWs + e b(ys , s) ds =: yt0 + yt1 , (4.5) ε −T ε −T t where α(t, s) = s a(u) du. We now define the space–time set B(h) mentioned above. If we write Var(yt0 ) = σ 2 v(t), then we find that v(t) solves the ODE ε v˙ = 2a(t)v + 1,

v(−T ) = 0.

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Let ξ(t) be a particular solution of this ODE with nonzero initial condition, given by ∫ 1 t 2α(t,s)/ε 1 ξ(t) = ξ(−T ) e2α(t,−T )/ε + e ds, ξ(−T ) = . ε −T 2|a(−T )|

2027

(4.6)

Then we define B(h) = {(y, t) : −T ⩽ t ⩽

 √ ε, |y| < h ξ(t)}

(4.7)

and the stopping time τB(h) = inf{t ⩾ −T : (yt , t) ̸∈ B(h)}. Before estimating τB(h) , we must first understand how a(t) and ξ(t) behave. Lemma 4.4. Let qtdet solve (4.1), define a(t) = t √ − 3(qtdet )2 and let ξ(t) (4.6). Then, √ be given by√ uniformly for x ∈ X , a(t) ≍ t for −T ⩽ t ⩽ − ε and |a(t)| = O( ε) for |t| ⩽ ε. We also have, uniformly for x ∈ X , ξ(t) ≍

1 √ |t| ∨ ε

and |ξ˙ (t)| = O(1/ε) for all −T ⩽ t ⩽

√

ε.

Proof. The assertions about a(t) follow from Lemma 4.3. We can use this to tell us how ξ(t) behaves, for which it is helpful to consider the case a(t) = t as an example. Furthermore, since ξ solves εξ˙ = 2a(t)ξ + 1, this tells us that |ξ˙ (t)| = O(1/ε).  Having established the behaviour of all relevant quantities, we can now prove the√following proposition telling us that sample paths are likely to remain in B(h) for all times t ⩽ ε. Proposition 4.5. There exists a constant C > 0 such that for all ε sufficiently small, all σ < h ≪ ε 3/4 and all initial conditions x ∈ X ,   √ h2 C P(τB(h) < ε) ⩽ 2 exp − 2 (1 − r (h, ε)) , ε 2σ √ where r (h, ε) = O( ε) + O(hε −3/4 ) uniformly for x ∈ X . √ Remark 4.6. Choosing h = kσ | ln  σ | with 0 large enough guarantees that the√ right-hand √ k >√ √ side tends to zero as σ ↓ 0 and that h ξ( ε) ≪ ε, in which case we may take q( ε) ≍ ε uniformly for all x ∈ X . √ Proof. Recall the decomposition yt = yt0 + yt1 from (4.5). We have for all t < τB(h) ∧ ε, ∫ 1 1 t α(t,u)/ε |y 1 | e |b(yu , u)| du √t ⩽ √ ξ(t) ξ(t) ε −T   ∫ Mh 2 1 t α(t,u)/ε ⩽ √ sup ξ(u) e du ε −T ξ(t) −T ⩽u⩽t c1 Mh 2 ε 3/4 for some constant c1 > 0, where we obtain the final inequality by bounding  √ ∫ 1 t α(t,u)/ε C/|t| √ for t ⩽ − √ε e du ⩽ C/ ε for |t| ⩽ ε ε −T ⩽

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and   √ C/ |t| for t ⩽ − ε √ sup ξ(u) ⩽ Cε −1/4 for |t| ⩽ ε. −T ⩽u⩽t √ 0 |/ ξ(t) < h(1 − c Mhε −3/4 ) for all −T ⩽ t ⩽ √ε then we must have Therefore, if |y 1 t √ τB(h) > ε. Letting H = h(1 − c1 Mhε −3/4 ), we obtain exactly as in the proof of the analogous Proposition 4.3 from [5] that for sufficiently small ε,     √ C |ys0 | H2 > H ⩽ 2 exp − 2 (1 − O( ε)) P sup √ √ ε 2σ ξ(t) −T ⩽t⩽ ε 1 √ ξ(t)





for some C > 0. Note we√ cannot apply that proposition directly because our function a(t) behaves differently for |t| ⩽ ε than the corresponding function there. In particular, here a(t) √ <0 for√t ≪ ε, whereas√ in [5] a(t) = t + O(t 2 ). However, in our case, |α(t, s)| = O((t − s) ε) for − ε ⩽ s ⩽ t ⩽ ε and we can check that this still allows us to suitably bound the term Pk appearing in Eq. (4.23) of their proof.  Step Two. We define for κ > 0 the space–time set √ K(κ) = {(q, t) : t ⩾ ε, q 2 ⩾ (1 − κ)t}.

(4.8) √ The boundary of K(κ) consists of the curves (± (1 − κ)t, t). For the present case, we only need to consider q ⩾ 0 (for the slow pulling regime √ in Section 4.2, we will also consider q < 0). √ Let t0 ⩾ √ ε and suppose that 0 < q(t0 ) < (1 − κ)t0 . Then for t > t0 and as long as 0 ⩽ qt ⩽ (1 − κ)t, we have qt ⩾ qtκ , where dqtκ =

1 σ κtqtκ dt + √ dWt , ε ε

q κ (t0 ) = q(t0 )/2.

(4.9)

Solving this SDE gives qtκ

1 σ 2 2 = q(t0 ) e(t −t0 )/2ε + √ 2 ε

∫

t

e(t

2 −s 2 )/2ε

dWs .

t0

We now state two lemmas that are analogues of Lemmas 3.2.10 and 3.2.11 from [6] and are proved in the same way. They tell √ us about the exit of qt from √ K(κ).√For the present section, we will only need to take t0 = ε and, by Step One, q0 = q( ε) ≍ ε. For the slow pulling regime, other initial conditions will be considered. Let τK(κ) = inf{t ⩾ t0 : (qt , t) ̸∈ K(κ)} and τκ0 = inf{t ⩾ t0 : qtκ ⩽ 0}. √ Lemma 4.7. Assume that qt starts at time t0 ⩾ ε in q0 > 0, where (q0 , t0 ) ∈ K(κ). Then there is C > 0, independent of t0 and q0 , such that for all t ⩾ t0 + ε/t0 , C √ √ −κ(t 2 −t 2 )/2ε 0 P(τK(κ) ⩾ t, τ0κ ⩾ t) ⩽ t0 t e . σ √ Lemma 4.8. Let qtκ start at time t0 ⩾ ε in q0 /2 > 0. Then there exist constants C1 , C2 > 0, independent of t0 and q0 , such that for all t ⩾ t0 , the probability of reaching zero before time t satisfies the bound   C2 q02 t0 C1 σ κ P(τ0 < t) ⩽ √ exp − . q 0 t0 σ2

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√ This second lemma shows that when q0 t0 ≫ σ , the linear process qtκ is unlikely to return to zero for any time t ⩾ t0 , while the first lemma shows that as t increases the probability of qt remaining in K(κ) decreases. Therefore, we have qt ⩾ qtκ until time τK(κ) and so qt must exit √ K(κ) through 2 > 0 such that for initial √ the curve (1 − κ)t. Indeed,√there are constants C1 , C√ time t0 = ε and any initial position q0 ≍ ε, we have for any t ⩾ 2 ε that   C1 ε 1/4 √ −κt 2 /(2ε) C1 σ C2 ε 3/2 κ P(τK(κ) < t, τ0 > t) ⩾ 1 − te − 3/4 exp − . (4.10) σ ε σ2 4/3 and so the third term on the right-hand side tends In the present fast pulling regime, √ ε≫σ to zero as σ ↓ 0. Picking t =√ 2kε| ln σ |, we see the second term also tends to zero as long as k > 1/κ. By a time of order ε| ln σ |, all paths will have left K(κ) through its upper boundary.

Step Three. Firstly, we will see how deterministic solutions behave when started from the ∗ (t) be the same solution of (4.2) that we considered in boundary of K(κ). For this, we let q+ the proof of Lemma 4.3, i.e. the unique real-valued solution √ √ existing for all times t ⩾ −T . For ∗ (t) ⩽ t > 0 and ε sufficiently small, we have t ⩽ q+ t + ε/t, which can easily be seen by the intermediate value theorem. For the sake of brevity, we shall write τ to mean τK(κ) . The following proposition tells us how qtdet,τ behaves, where qtdet,τ solves q˙tdet,τ =

1 det,τ (tq − (qtdet,τ )3 + ε), ε t

qτdet,τ =



(1 − κ)τ .

(4.11)

Proposition 4.9. Assume that κ ∈ (1/2, 2/3) and let η = 2 − 3κ > 0. There is a constant C > 0 such that the solution, qtdet,τ , of (4.11) satisfies   ε 2 2 ∗ 0 ⩽ q+ (t) − qtdet,τ ⩽ C 3/2 + (q ∗ (τ ) − q det,τ (τ )) e−η(t −τ )/2ε t for all t ⩾ τ and ε sufficiently small. Remark 4.10. The condition κ > 1/2 guarantees that paths do not re-enter K(κ) after leaving, while κ < 2/3 ensures that the potential is convex outside of K(κ). ∗ (t) follows since q det,τ (τ ) < q ∗ (τ ) and Proof. The inequality qtdet,τ ⩽ q+ + ∗ ′ (q+ ) (t) =

∗ (t) q+ ∗ 3q+ (t)2 − t

> 0.

(4.12)

The proof of the other inequality follows along the same lines as that given for the analogous Proposition 4.11 from [5]. Note, however, that unlike there we only need to take ε sufficiently small and not t. This is because in our case the value of a0∗ , which is defined in Eq. (4.99) ∗ of [5], is given√by −2(1 + oε (1)), √ rather than −2(1 √ + ot (1)). Similarly, M = 3(1 + oε (1)). ∗ As q+ (t) ⩽ t + ε/t ⩽ (3/2) t for t ⩾ ε and ε small, we can use (4.12) to show ∗ )′ (t) ⩽ (3/4)t −1/2 , giving K ∗ = 3/4, where K ∗ is also defined in (4.99). In [5], K ∗ = 1/2, (q+ but the proof just requires that K ∗ < 1.  Now that we understand how qtdet,τ behaves, the final step is to show that qt , starting at the same point, stays close. Having shown that the analogue of Proposition 4.11 from [5] holds, the proofs of the subsequent bounds there can easily be extended to our case and we now show what

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these are. Let ξ τ (t) =

1 1 τ e2α (t,τ )/ε + 2|a τ (τ )| ε

∫ τ

t

e2α

τ (t,s)/ε

ds,

where a τ (τ ) = t − 3(qtdet,τ )2 is the linearisation of the drift term around qtdet,τ and α τ (t, s) = t s a(u) du. As is shown in Lemma 4.12 from [5], it follows from Proposition 4.9 above that |a τ (τ )| ≍ t so that ξ τ (t) ≍ 1/t. Now we write  Aτ (h) = {(q, t) : t ⩾ τ, |q − qtdet,τ | ⩽ h ξ τ (t)} (4.13) and let τAτ (h) = inf{t ⩾ τ : (qt , t) ̸∈ Aτ (h)}. The following bound on τAτ (h) can be proved in the same way as Theorem 2.12 in [5]. It tells us that for κ ∈ (1/2, 2/3) and any t2 > 0, there exist constants C, h 0 > 0 such that for h < h 0 τ and ε sufficiently small,  [  ] √ C 1 h2 h τ, (1−κ)τ P (τAτ (h) < t2 ) ⩽ 2 exp − 1 − O(ε) − O . (4.14) 2 2 τ ε σ √ The right-hand side √ becomes small by choosing h = kσ | ln σ | for k large enough, for which we note that τ ⩾ ε by definition so that h ≪ τ . √ Step √Four. Let us suppose that t2 ⩾ 1. By Step Three, we may write q(t2 ) = t2 + O(σ | ln σ |) + O(ε) independently of τK(κ) . For a given q(t2 ), let qtdet be the corresponding deterministic solution starting at the same point. We can again √ obtain a similar bound√as in Proposition 4.9, but for simplicity let us just say that (9/10) t ⩽ qtdet ⩽ (11/10) t for all t ⩾ 1. Letting yt = qt − qtdet , we define τ (γ ) = inf{t ⩾ t2 : |yt | > t 1/2−γ } for 0 < γ < 1/2. We again decompose yt into a linear part, yt0 , and nonlinear part, yt1 , as in (4.5). Then a(t) = t − 3(qtdet )2 ≍ −t uniformly√for all t ⩾ 1 and the function b(y, t) containing the nonlinear terms now satisfies |b(yt , t)| < M t yt2 for all t < τ (γ ) and some constant M > 0 independent of t2 . We will show that P(τ (γ ) < ∞) → 0. For t ⩽ τ (γ ), we have ∫ 1 t α(t,s)/ε |yt1 | ⩽ e |b(yu , u)| du ε t2 ∫ Mt 3/2−2γ t α(t,s)/ε ⩽ e du ε t2 < Ct 1/2−2γ , where the final inequality holds uniformly in t and the constant C > 0 is independent of t2 . Therefore, if |yt0 | < H (t) for all t ⩾ t2 , where H (t) = t 1/2−γ (1 − Ct −γ ), then we must have τ (γ ) = ∞. Note that 1 − Ct −γ > 0 and H˙ (t) > 0 for all t ⩾ t2 by taking t2 large enough. We have     ∞ − |yt0 | |yt0 | P sup >1 ⩽ P sup >1 , t⩾t2 H (t) s j ⩽t⩽s j+1 H (t) j=0 where t2 = s0 < s1 < · · · is chosen by −α(s j+1 , s j ) = ε 2 . Note that, uniformly in j, we have s 2j+1 − s02 ≍ −α(s j+1 , s0 ) = −α(s j+1 , s j ) − · · · − α(s1 , s0 ) = ( j + 1)ε2 , which shows that s j → ∞ as j → ∞. Call the summand on the right-hand side above P j . As H (t) is increasing, we can further bound P j by replacing H (t) with H (s j ). We can also use for s j ⩽ t ⩽ s j+1 the

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inequality |yt0 |

 ∫  σ  = √ ε

t

α(t,s)/ε

e s0

  dWs  ⩽ eα(s j )/ε

  ∫ t  σ  −α(s)/ε √ e dWs  .  ε s0

This gives for all j ⩾ 0,    √ ∫ t   ε −α(s)/ε −α(s )/ε j e dWs  > Pj ⩽ P sup  e H (s j ) σ s0 ⩽t⩽s j+1 s0   ε e2α(s j+1 ,s j )/ε H (s j )2 s ⩽ 2 exp − 2σ 2 s0j+1 e2α(s j+1 ,s)/ε ds   c1 s j H (s j )2 ⩽ 2 exp − , 2σ 2 where the constant c1 > 0 in the final inequality is independent of j. Note that the second inequality comes from Lemma B.1.3 in the Appendix of [6], which states that     ∫ s  2   δ P sup  ϕ(u) dWu  ⩾ δ ⩽ 2 exp −  t 2 0 ϕ(u)2 du 0⩽s⩽t 0 for all Borel-measurable deterministic functions ϕ : [0, t] → R, and the final inequality uses that α(s j+1 , s) ≍ −(s 2j+1 − s 2 ) uniformly for all s0 ⩽ s ⩽ s j+1 and all j. Summing over j ⩾ 1 and using that s j − s j−1 ⩾ Cε 2 /s j uniformly in j, we have ∞ −

Pj =

j=1

∞ −

Pj

j=1

s j − s j−1 s j − s j−1

∞ C − P j s j (s j − s j−1 ) ε 2 j=1 ∫ ∞ ⩽C s P(s) ds,

⩽

t2

where c1 s H (s)2 P(s) = 2 exp − 2σ 2 



c2 s 2(1−γ ) ⩽ 2 exp − σ2 



and c2 > 0 is a constant. We have used that s P(s) is decreasing when bounding the series by the integral above. Then   ∫ ∞ 2(1−γ ) c t 2 s P(s) ds ⩽ Cσ 2 exp − 2 2 σ t2 for some constant C > 0 depending on t2 and γ , so that     2(1−γ ) c2 t2 |yt0 | Cσ 2 P sup > 1 ⩽ P0 + 2 exp − ε σ2 t⩾t2 H (t) and the right-hand side tends to zero as σ ↓ 0.

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4.2. Slow pulling We now consider the slow pulling regime from Proposition 4.2. In this case, the noise dominates the dynamics and cancels out the asymmetry caused by pulling. The process qt should, therefore, behave similarly to q˜t , where dq˜t =

1 σ (t q˜t − q˜t3 ) dt + √ dWt , ε ε

q(−T ˜ ) = 0.

(4.15)

As we have chosen q(−T ˜ ) = 0, the law of q˜ is entirely symmetric about zero. The strategy is as follows: √ (1) Recall from [5,6] that q˜t stays close√to the origin with high probability. At time t = ε, its typical spreading is of order σ ε−1/4 | ln σ |. (2) Show that paths of qt stay close to those of q˜t until q˜t leaves the diffusion-dominated strip S(h) defined below. (3) Show that qt then exits the slightly larger strip, K(κ), without returning to the origin. (4) Show that qt then finally falls into the potential well on the same side as it left K(κ) and remains there. Step One. This step is the same as Step One from the previous section, except now we are analysing the behaviour of q˜t . Unlike in the previous section, we can use directly the results of [5,6], which we now summarise. We again define the function ξ(t) as in the last section and now a(t) := t − 3(q˜tdet )2 , where 1 det q˙˜ t = (t q˜tdet − (q˜tdet )3 ), ε

det q˜−T = 0.

Clearly, q˜ det ≡ 0 and so now a(t) ≡ t. Again, ξ(t) ≍ 1/(|t| ∧ the space–time domain  √ B(h) = {(q, ˜ t) : −T ⩽ t ⩽ ε, |q| ˜ < h ξ(t)}

√ √ ε) for −T ⩽ t ⩽ ε. We define

and the stopping time τB(h) = inf{t ⩾ −T : (q˜t , t) ̸∈ B(h)}. According to Theorem 2.10 √ from [5], there are constants C, h 0 > 0 such that for ε sufficiently small and h ⩽ h 0 ε, [  2 ]  √ √ C h h2 P(τB(h) < ε) ⩽ 2 exp − 2 1 − O( ε) − O . ε ε 2σ √ √ Choosing h = k√large enough, the right-hand side tends to zero. At time ε, we √ kσ | ln σ | for may take |q( ˜ ε)| = O(σ ε−1/4 | ln σ |). √ √ √ Step Two. In the fast pulling section, we saw that at time ε, q( ε) ≍ ε, from which we could then show its subsequent exit from K(κ). Before looking in this slow pulling section at the exit of qt from K(κ), we must √ first recall a result from [5] about the exit of q˜t from a smaller strip S(h), defined for t ⩾ ε as   √ h S(h) = (q, ˜ t) : t ⩾ ε, |q| ˜ <√ . t √ We shall also use the stopping time τS (h) = inf{t ⩾ ε : (q˜t , t) ̸∈ S(h)}. For h ≈ σ , the set S(h) is a strip around the origin in which the drift term acting on q˜t is very small, so that its diffusion term dominates.

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√ √ √ Let h ∗ := h 0 σ | ln σ |, where h 0 > 0 is a constant sufficiently large so that (q( ˜ ε), ε) ∈ S(h ∗ ) almost surely as σ ↓ 0. Applying Proposition 4.7 from [5] with the choices h = h ∗ and µ = 2, we√ see that there exists C > 0 such that for all σ sufficiently small and all initial conditions (q0 , ε) ∈ S(h ∗ ),  ∗ 2   h 2 (t 2 − ε) ∗ ∗ P(τS (h ) ⩾ t) ⩽ exp − [1 − O(1/ ln(h /σ ))] , σ 3 2ε √ as long as σ | ln σ |3/2 = √ O( ε), which we already assume in the slow pulling regime. We can check that by taking t = 2kε ln(h ∗ /σ ) with√ k > 0 sufficiently large, the right-hand side tends to zero. For such a choice of k, we define t ∗ = 2kε ln(h ∗ /σ ) and henceforth assume τS (h ∗ ) ⩽ t ∗ . The important point here is that by symmetry, q˜t exits S(h ∗ ) through either boundary with equal probability. Now that we understand the behaviour of q˜t up until its exit from S(h ∗ ), we turn to qt . The following lemma shows that qt is close to q˜t at time τS (h ∗ ) . Lemma 4.11. Let qt solve (2.4) with any initial condition q(−T ) = x ∈ X , and suppose ε(σ ) ≪ σ 4/3 | ln σ |−13/6 . Then for almost all paths in the set √ √ {(q( ˜ ε), ε) ∈ S(h ∗ ), τS (h ∗ ) ⩽ t ∗ }, we have    3/4 ε | ln σ |13/8 qτS(h ∗ ) = q˜τS(h ∗ ) 1 + O . σ The proof of this lemma is based on the following simple comparison of qt and q˜t . Lemma 4.12. Let qt solve (2.4) with initial condition q(−T ) = x ∈ X and let q˜t solve (4.15). We have, almost surely, for all t ⩾ −T , ∫ t 2 2 2 2 q˜t + x e(t −T )/2ε ⩽ qt ⩽ q˜t + e(t −s )/2ε ds (4.16) −T

if x ⩽ 0 and t

∫

e(t

q˜t ⩽ qt ⩽ q˜t +

2 −s 2 )/2ε

ds + x e(t

2 −T 2 )/2ε

(4.17)

−T

if x ⩾ 0. Proof. We have qt = x e(t

2 −T 2 )/2ε

∫

t

+

e(t

2 −s 2 )/2ε

−T

σ +√ ε

∫

t

e(t

2 −s 2 )/2ε

ds −

1 ε

∫

t

e(t

2 −s 2 )/2ε

−T

dWs .

(4.18)

−T

By the comparison principle, qt ⩾ qˆt almost surely, where qˆt solves dqˆt =

1 σ (t qˆt − qˆt3 ) dt + √ dWt , ε ε

qs3 ds

q(−T ˆ ) = x.

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Using this lower bound for q in the second integral on the right-hand side of (4.18) gives ∫ t 2 2 qt ⩽ qˆt + e(t −s )/2ε ds. −T

When x ⩽ 0, we have qˆt ⩽ q˜t almost surely, which gives the upper bound in (4.16). For the lower bound, we have ∫ ∫ t σ 1 t (t 2 −s 2 )/2ε 3 2 2 (t 2 −T 2 )/2ε e qˆs ds + √ qˆt = x e − e(t −s )/2ε dWs ε −T ε −T ∫ t ∫ t 1 σ 2 2 2 2 2 2 ⩾ x e(t −T )/2ε − e(t −s )/2ε q˜s3 ds + √ e(t −s )/2ε dWs ε −T ε −T = x e(t

2 −T 2 )/2ε

+ q˜t .

The case x ⩾ 0 is easier and does not involve qˆt . It follows along similar lines. √ Proof of Lemma 4.11. For all ε ⩽ t ⩽ t ∗ , we have   ∫ t √ h∗ 4 √ (t 2 −s 2 )/2ε e ds ⩽ C1 ε ⩽ C2 ε| ln σ |2 σ −T



(4.19)

and for all x ∈ X , |x| e(t

2 −T 2 )/2ε

 ⩽ C1

h∗ σ

4

e−T

2 /2ε

⩽ C2 | ln σ |2 e−T

2 /2ε

.

Of these two estimates, (4.19) gives the larger upper bound. Next observe that τS (h ∗ ) ⩽ t ∗ ⩽ Cε 1/2 | ln σ |1/4 . Therefore, √

1 |q˜τS(h ∗ ) |

=

τS (h ∗ ) h∗

⩽C

ε 1/4 . σ | ln σ |3/8

Using (4.19) and the above inequality together with Lemma 4.12 gives the result.



∗ Step Three. Now we analyse the behaviour in the √ of qt , rather than q˜t , for t > τS (h ) . As we saw κ as long fast pulling case, if q starts at time t ⩾ ε at q > 0 with (q , t ) ∈ K(κ) then q ⩾ q t t 0 0 0 0 t √ as 0 < qt < (1 − κ)t, where qtκ was defined in (4.9). Unlike in the previous section, we now have to also consider negative initial conditions. We would like a bound of the form qt ⩽ qtκ in such cases, but now the bias of qt in the positive direction makes this more difficult. In order to obtain a corresponding comparison with qtκ , we need an additional assumption on t0 and q0 as set out in the following lemma.

√ Lemma 4.13. Suppose that at time t0 ⩾ ε, qt√starts at q0 < 0, where (q0 , t0 ) ∈ K(κ) and |q0 | ≫ ε/t0 . Then we have qt ⩽ qtκ as long as − (1 − κ)t ⩽ qt ⩽ 0 and ε is sufficiently small. √ Proof. For − (1 − κ)t0 < q0 < 0, it is certainly true by comparison of the drift and initial conditions that qt is bounded above by solutions of dz tκ =

1 σ (κt z tκ + ε) dt + √ dWt , ε ε

z tκ0 = q0 ,

M. Allman et al. / Stochastic Processes and their Applications 121 (2011) 2014–2042

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√ as long as − (1 − κ)t ⩽ qt ⩽ 0. The result follows since ∫ t ∫ t σ 2 2 κ κ(t 2 −t02 )/2ε κ(t 2 −s 2 )/2ε z t = q0 e eκ(t −s )/2ε dWs + e ds + √ ε t0 t0 ∫ t σ 2 2 2 2 ⩽ (q0 + Cε/t0 ) eκ(t −t0 )/2ε + √ eκ(t −s )/2ε dWs ε t0 ∫ t q0 κ(t 2 −t 2 )/2ε σ 2 2 0 eκ(t −s )/2ε dWs ⩽ e +√ 2 ε t0 κ = qt .  This lemma shows that, under suitable conditions, we may compare qt and qtκ for both positive and negative initial conditions q0 . For q0 < 0 we get analogous bounds Lemmas 4.7 √ to those in √ and 4.8. In the previous section, we applied those lemmas with t0 = ε and q0 ≍ ε. We now √ apply these lemmas with t0 = τS (h ∗ ) and |q0 | ≍ h ∗ / τS (h ∗ ) (see Lemma 4.11). Note that if √ ε ⩽ τS (h ∗ ) ⩽ t ∗ then ε/t0 ≪ |q0 | and so the conditions of Lemma 4.13 are√satisfied. We obtain a bound similar to (4.10) and again see that K(κ) is left by a time of order ε| ln σ |. Step Four. When qt exits K(κ) on the positive side, this part is exactly the same as Step Three from the previous section. The other case when qt exits K(κ) on the negative side is √ similar. ∗ (t), another real-valued solution of (4.2) existing for t ⩾ Firstly, we introduce q− ε and √ √ ∗ (t) ⩽ − t + ε/t and (q ∗ )′ (t) < 0 for all such t and ε satisfying the bounds − t ⩽ q− − ∗ ∗ small. Note there is also a third real-valued solution of (4.2) between √ q− and q+ , which is√an ∗ unstable equilibrium branch. By taking ε small, we have q− (t) < − (1 − κ)t for all t ⩾ ε. det,τ , of (4.11), with Again writing τ = τK(κ) , we need √ to check that the deterministic solution, qt det,τ initial condition q (τ ) = − (1 − κ)τ , satisfies a bound as in Proposition 4.9 of the form  ε  2 2 ∗ 0 ⩽ qtdet,τ − q− (t) ⩽ C 3/2 + (q det,τ (τ ) − q ∗ (τ )) e−η(t −τ )/2ε t for all t ⩾ τ . The main thing to ensure is that qtdet,τ , which has a bias in the positive direction, does not re-enter the set K(κ) after having left. To see that √ that √ this is indeed the case, first note the derivative with respect to t of the boundary curve, − (1 − κ)t, is given by − 21 t −1/2 1 − κ. √ The derivative of qtdet,τ when on the boundary of K(κ) is given by 1ε t 3/2 (−κ + ε) 1 − κ. Using √ that t ⩾ ε, we see that the inequality √ √ 1 1 3/2 t (−κ + ε) 1 − κ < − t −1/2 1 − κ ε 2 holds when κ > 1/2 + ε, which is true for all κ > 1/2 by taking ε sufficiently small. Having √ ∗ (t) ⩽ q det,τ ⩽ − (1 − κ)t for all t ⩾ τ and ε sufficiently small, the rest of established that q− t the proof follows like that of the analogous Proposition 4.11 from [5]. The subsequent estimate (4.14) above showing the concentration of qt in the set Aτ (h) then follows, where Aτ (h) was defined in (4.13). Finally,√we can show as in Step Four from the fast pulling section that qt stays in a neighbourhood of − t for all t ⩾ t2 . 5. The full solution We now consider the full equation (2.2) and show that for sufficiently small mass (large β), it behaves like the overdamped solution of the previous section. The general strategy is as in

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Section 4, namely we first show that if we start the system at the origin at time s ≪ −1, then the solution at time −1 belongs to a suitable set. This is done in the following two propositions. We then provide a result that is uniform over all solutions starting from the set in question. The first step is achieved by the following statement. Proposition 5.1. Let Xˆ = [−ε 1/2−β , ε1/2−β ], Vˆ = [−ε − and let qt solve (2.2) with β > 1/2. Then, we have

1+3β 2

, ε−

1+3β 2

], let T ⩾ 1 be a constant,

ˆ = 1. lim lim inf Ps (q−2T ∈ Xˆ , p−2T ∈ V)

σ,ε→0 s→−∞

Proof. We fix some arbitrary starting time s < −2T and we consider the solution to (2.2) with initial condition qs = ps = 0. We define the function Ψ ( p, q, t) by εβ 2 t 1 1 p − q 2 + q 4 − q + pq. 2 2ε 4ε 2 Applying Itˆo’s formula to Ψ ( pt , qt , t), we obtain   1 1 1 dΨ ( pt , qt , t) ⩽ −ε −β Ψ ( pt , qt , t) + σ 2 ε −1−β − qt2 − β qt dt + dM(t), 2 2ε 2ε Ψ ( p, q, t) =

where M is some continuous martingale. Using the inequality ε β−1 q 2 + q ⩾ −ε 1−β /4, we see that d 1 1 EΨ ( pt , qt , t) ⩽ −ε −β EΨ ( pt , qt , t) + σ 2 ε −1−β + ε 1−2β . dt 2 8 It follows immediately that EΨ ( pt , qt , t) ⩽ σ 2 /(2ε) + ε 1−β /8. For t ⩽ −2T ⩽ −2, we have Ψ ( p, q, t) ⩾

εβ 2 1 2 εβ 2 1 1 1 1 p + q + q4 − β q2 − q ⩾ p + q 2 − ε − ε 1−2β , 4 ε 4ε 4ε 4 2ε 16

where we used the inequality 2 pq ⩾ −ε β p 2 − ε −β q 2 , which tells us that E(qt2 ) ⩽ 2ε 2 + ε 2−2β /8 + σ 2 + ε 2−β /4, E( pt2 ) ⩽ 4ε 1−β + ε 1−3β /4 + 2σ 2 ε −1−β + ε 1−2β /2. The result then follows from Chebyshev’s inequality, noting that β > 1/2.



ˆ vˆ Now we use the previous proposition to restart the process at time −2T . We denote by Px, ˆ p−2T = v. ˆ the law of the solution of (2.2) starting at time −2T with q−2T = x,

Proposition 5.2. Let X = [−1, 1], V = [−ε −β , ε−β ], let T ⩾ 1 be a constant, and let qt solve (2.2) with β > 2. Then, we have lim

inf

σ,ε→0 x∈ ˆ Xˆ ,v∈ ˆ Vˆ

ˆ vˆ Px, (q−T ∈ X , p−T ∈ V) = 1.

Proof. Let 1 2 εβ ε 2β 2 ε β−1 4 q + pq + p + q . 4 2 2 4 Applying Itˆo’s formula to Ψ ( pt , qt ), we obtain Ψ ( p, q) =

M. Allman et al. / Stochastic Processes and their Applications 121 (2011) 2014–2042

dΨ ( pt , qt ) =

2037

 β+1 1 ε t 1 ε − pt2 + qt2 − qt4 + qt ε 2 2 2 2  2 σ dt + dM(t), + ε β t pt qt + ε β+1 pt + 2

where M is some continuous martingale. Using the inequality 2 pq ⩽ ε 2 p 2 + ε −2 q 2 , and that t ∈ [−2T, −T ] and β > 2, we have for sufficiently small ε,   1 1 ε2 σ2 dΨ ( pt , qt ) ⩽ − Ψ ( pt , qt ) + 4ε β+1 + + dt + dM(t). ε 2 2 2 It follows that EΨ ( pt , qt ) ⩽ e−(t+2T )/2ε EΨ ( p−2T , q−2T ) + σ 2 + 8ε β+1 + ε 2 . We can then use the bounds Ψ ( p, q) ⩾ ε 2β p 2 /4 and Ψ ( p, q) ⩾ q 2 /8, along with Chebyshev’s inequality, to obtain the result.  We would like to use a singular perturbation approach to show that for t ⩾ −T and suitably large β, sample paths of qt can be approximated by those of an overdamped equation starting at −T . For this, we need to first consider a process that is similar to qt but has better regularity. To this end, we introduce the processes Q and P that solve dQ t = Pt dt,

Q −T = P−T = 0, σ ε β dPt = −Pt dt + √ dWt . ε

(5.1)

We find that Pt = σ ε−1/2−β

∫

t

e−(t−s)ε

−β

dWs

−T

and σ Q t = √ Wt − ε β Pt = σ ε−1/2−β ε

∫

t

e−(t−s)ε

−β

W (s) ds.

−T

For δ > 0 we now define two events, E 1 and E 2 , by E 1 = {|Q t | > σ ε−1/2−δ for some t ∈ [−T, t2 ]}, E 2 = {|Pt | > σ ε−1/2−β/2−δ for some t ∈ [−T, t2 ]}. Lemma 5.3. For all δ > 0, all T ⩾ 1 and all t2 > 0,   ε −2δ P(E 1 ∪ E 2 ) ⩽ 8(t2 + T ) exp − 2(t2 + T ) holds for ε > 0 sufficiently small. t −β Proof. Since Q t = σ ε−1/2−β −T e−(t−s)ε W (s) ds, it follows that P(E 1 ) ⩽ P(|Wt | > ε−δ for some t ∈ [−T, t2 ])   ε −2δ . ⩽ 4 exp − 2(t2 + T )

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M. Allman et al. / Stochastic Processes and their Applications 121 (2011) 2014–2042

We also find that there exists C > 0, independent of δ and t2 , such that for ε sufficiently small,   1 −2δ −2δ−1 . P(E 2 ) ⩽ C(t2 + T )ε exp − ε 2 The result follows by combining these two estimates and taking ε small.



Now let yt = qt − Q t . It solves, almost surely, the second-order ODE 1 ε β y¨ = − y˙ + (t (yt + Q t ) − (yt + Q t )3 + ε), ε

y(−T ) = x,

y˙ (−T ) = v.

The following proposition shows that for almost all paths in (E 1 ∪ E 2 )c , y may be approximated by the solution of a first-order ODE. Note that the condition on β is a little stronger than necessary, but is required later on in this section. Proposition 5.4. For all t2 > 1, there exists C = C(t2 ) > 0 such that for all β > 2, all 0 < δ < β/2 − 1, all σ 2/(1+2δ) ≪ ε ≪ 1 and almost all paths in (E 1 ∪ E 2 )c ,      y˙ − 1 (t (yt + Q t ) − (yt + Q t )3 + ε) ⩽ C max{ε β−2−δ , σ ε−3/2+β/2−2δ }   ε for all t ∈ [−T + 2ε β−δ , t2 ] and σ sufficiently small. Proof. If we write z = y˙ , then almost surely the pair (y, z), which is differentiable, solves y˙ = z, 1 ε β z˙ = −z + g(t, yt + Q t ), ε

√ where g(t, yt + Q t ) = t (yt + Q t ) − (yt + Q t )3 + ε. For t > 0, we have √ g(t, t) ≈ 0 so that we do not expect yt √ + Q t , or indeed yt , to be much larger than t. Therefore, we let τ = inf{t ⩾ −T : |yt | > 2 t2 }. On (E 1 ∪ E 2 )c , there is C > 0 depending on t2 such that for all −T ⩽ t ⩽ τ ∧ t2 , |g(t, yt + Q t )| < C. We solve the equation for z to give ∫ t −β −β e−(t−s)ε g(s, ys + Q s ) ds (5.2) z t = v e−(t+T )ε + ε −(1+β) −T

almost surely, from which we deduce that |z t | ⩽ ε −β e−(t+T )ε + C/ε for all t ⩽ τ ∧ t2 . This immediately shows that for −T ⩽ t ⩽ (−T + 2ε β−δ ) ∧ τ and sufficiently small ε, −β

|yt − x| ⩽ 1 + Cε β−δ−1

(5.3)

and so τ > −T + 2ε β−δ . For −T + ε β−δ ⩽ t ⩽ τ ∧ t2 (note we have written ε β−δ , not 2ε β−δ ), we have |z t | < C/ε. Furthermore, for such t we find   d   g(t, yt + Q t ) ⩽ C max{ε −1 , σ ε−1/2−β/2−δ }.  dt  Now we apply the Laplace method to the integral in (5.2). For t ⩾ −T + 2ε β−δ , decompose the integral as ∫ t ∫ t−εβ−δ ∫ t = + . −T

−T

t−ε β−δ

M. Allman et al. / Stochastic Processes and their Applications 121 (2011) 2014–2042

2039

Then, by the boundedness of g, we have  ∫   t−εβ−δ −δ −β   e−(t−s)ε g(s, ys + Q s ) ds  < C e−ε .    −T For the remaining integral, we use a Taylor expansion of g to give g(s, ys + Q s ) ⩽ g(t, yt + Q t ) + C(t − s) max{ε −1 , σ ε−1/2−β/2−δ }. Then ∫

t

t−ε β−δ β

e−(t−s)ε

−β

g(s, ys + Q s ) ds

⩽ ε g(t, yt + Q t ) + Cε β e−ε

−δ

+ C max{ε 2β−1−δ , σ ε−1/2+3β/2−2δ },

which tells us that for −T + 2εβ−δ ⩽ t ⩽ τ ∧ t2 and σ sufficiently small, zt ⩽

1 g(t, yt + Q t ) + C max{ε β−2−δ , σ ε−3/2+β/2−2δ }. ε

(5.4)

In a similar way, we can also show that zt ⩾

1 g(t, yt + Q t ) − C max{ε β−2−δ , σ ε−3/2+β/2−2δ }. ε

(5.5)

We will now show √ that the assumption τ ⩽ t2 leads to a contradiction. For this, we will show that if yτ = +2 t2 , then the right-hand side √ of (5.4) is strictly negative, whereas we should √ have z τ ⩾ 0 by continuity. The case yτ = −2 t2 is similar. First, we note that if yτ = +2 t2 , then there are constants C1 , C2 > 0 depending on t2 such that g(τ, yτ + Q τ ) ⩽ −C1 + C2 σ ε−1/2−δ and for σ sufficiently small, the right-hand side is strictly negative and bounded away from zero. Then the conditions on ε, β and δ guarantee that the right-hand side of (5.4) is strictly negative. This means that we must have τ > t2 and so (5.4) and (5.5) hold for all −T + 2ε β−δ ⩽ t ⩽ t2 , from which the result follows.  Now we will use Proposition 5.4 to tell us something about the SDE (2.2). Proposition 5.5. For all t2 > 1, there exists C = C(t2 ) > 0 such that for all β > 2, all 0 < δ < β/2 − 1, all σ 2/(1+2δ) ≪ ε ≪ 1 and almost all paths in (E 1 ∪ E 2 )c , qt− ⩽ qt + ε β Pt ⩽ qt+ for all t ∈ [−T + 2ε β−δ , t2 ] and σ sufficiently small, where dqt± =

1 ± σ [tqt − (qt± )3 + ε(1 ± r (σ ))] dt + √ dWt , ε ε

q (−T + 2ε ±

β−δ

) = x ± 3,

with Wt the same Brownian motion appearing in (2.2) and (5.1), and r (σ ) = C max{ε β−2−δ , σ ε−3/2+β/2−2δ }.

(5.6)

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Proof of Theorem 5.5. We will show the upper bound. The lower bound is similar. Letting t0 = −T + 2ε β−δ , we know by Proposition 5.4 that there exists C > 0 and a process yt+ solving y˙t+ =

1 g(t, yt+ + Q t ) + C max{ε β−2−δ , σ ε−3/2+β/2−2δ }, ε

yt+0 = x + 2,

such that yt ⩽ yt+ for all t ∈ [t0 , t2 ], where yt = qt − Q t . Note the initial condition for yt+ is + chosen using (5.3). √ Therefore, qt ⩽ yt + Q t . In the same way +as in Proposition 5.4, we can + show |yt | < 4 t2 for all t0 ⩽ t ⩽ t2 by considering the sign of y˙t . Now define ηt := yt+ + Q t + ε β Pt and note that for all t0 ⩽ t ⩽ t2 , |ηt | < C for some constant C > 0 depending on t2 . It solves dηt =

1 [t (ηt − ε β Pt ) − (ηt − ε β Pt )3 + ε + C max{ε β−1−δ , σ ε−1/2+β/2−2δ }] dt ε σ + √ dWt ε

with initial position ηt0 ⩽ x + 3. Denote the drift term above by f (t, ηt , Pt , ε). We will now show that f is bounded in such a way that allows us to use a comparison principle. As we are working on (E 1 ∪ E 2 )c , we know that |ε β Pt | ⩽ σ ε−1/2+β/2−δ for all t ∈ [t0 , t2 ] by definition. Note also that by the conditions on δ and ε, max{ε β−1−δ , σ ε−1/2+β/2−2δ } ≪ ε. We then have f (t, ηt , Pt , ε) ⩽

1 [tηt − ηt3 + ε(1 + r (σ ))], ε

where r (σ ) = C max{ε β−2−δ , σ ε−3/2+β/2−2δ } for some constant C > 0 depending on t2 . Let qt+ be the solution of dqt+ =

1 + σ [tqt − (qt+ )3 + ε(1 + r (σ ))] dt + √ dWt , ε ε

qt+0 = x + 3.

By Lemma A.1 below, ηt ⩽ qt+ for all t ∈ [t0 , t2 ] and almost all paths in (E 1 ∪ E 2 )c . Therefore, qt + ε β Pt ⩽ ηt ⩽ qt+ , which gives the upper bound.  This proposition allows us to complete the proof of Theorem 2.3. Proof of Theorem 2.3. Firstly, let us consider the fast pulling case. That is, σ 4/3 | ln σ |2/3 ≪ ε ≪ 1. In this case, when β > 2 the term 1 − r (σ ) appearing in (5.6) is strictly positive and bounded away from zero for all σ sufficiently small. Then the analysis of Section 4.1 can be applied to qt− (the slightly different drift term does not matter). By Lemma 5.3, we can assume −1/2+β/2−δ for all t ∈ [−T +2ε β−δ , t ]. Therefore, (E 1 ∪ E 2 )c to hold, in which √ case |ε β Pt | ⩽ 2 √σε we find that qt− − ε β Pt > γ t for all c1 ε| ln σ | ⩽ t ⩽ t2 , where c1 , γ > 0 are suitably chosen constants. For the slow pulling case, let 0 < δ < β/2 − 1 and σ 2/(1+2δ) ≪ ε ≪ σ 4/3 | ln σ |−13/6 . Then again 1±r (σ ) is positive and bounded away from zero for σ small and the analysis in Section 4.2 applies to qt+ and qt− . Note that, in the limit, qt− and qt+ must “go the same way” by comparison of their drift terms and initial positions and we know by Proposition 4.2 that the probabilities are 1/2 in either direction. As above, the term ε β Pt does not change anything. Therefore, qt behaves in the same way as qt− and qt+ . 

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Acknowledgements M.A. would like to thank Barbara Gentz for many helpful discussions during a visit to Bielefeld University, supported by SFB 701, and the Courant Institute for its hospitality. M.A. is supported by EPSRC Award EP/P502810/1 and a Warwick University ASSEC grant. V.B. is supported by EPSRC Fellowship EP/D07181X/1. M.H. gratefully acknowledges support by the EPSRC through an Advanced Research Fellowship EP/D071593/1 and by the Leverhulme Trust through a Philip Leverhulme Prize. Appendix. A comparison principle for SDE We present a lemma that is a slightly modified version of Theorems 5.1 and 5.2 appearing in [3]. Our proof follows those given there. Let (Ω , F, P) be a probability space on which is defined a one-dimensional Brownian motion W adapted to a filtration (Ft )t⩾0 . For t ⩾ 0, let X (t) and Y (t) be two real-valued processes evolving according to ∫ t X (t) = X (0) + a(s, X (s), Z (s)) ds + C Wt , 0 ∫ t c(s, Y (s)) ds + C Wt , Y (t) = Y (0) + 0

where C is a constant, a : [0, ∞)×R2 → R, c : [0, ∞)×R → R are continuous functions, Z (t) is an Ft -adapted process with continuous sample paths almost surely and Z (0) = z 0 , which is F0 -adapted. We assume that X (0) ⩽ Y (0) almost surely, where X (0) and Y (0) are F0 -adapted. Lemma A.1. Suppose there are constants C1 , C2 > 0 such that whenever |x| < C1 and |z| < C2 , a(t, x, z) < c(t, x) for all t ⩾ 0. If, almost surely, |X (t)| < C1 and |Z (t)| < C2 for all t ⩾ 0 then P(Y (t) ⩾ X (t) for all t ⩾ 0) = 1. Proof. Define τ = inf{t > 0 : Y (t) − X (t) < 0} and set τ = +∞ if Y (t) ⩾ X (t) for all t ⩾ 0. This is a stopping time because if t > 0 then  {τ ⩾ t} = {Y (r ) − X (r ) ⩾ 0} ∈ Ft . r ∈[0,t]∩Q

Put D = {τ < +∞} and assume that P(D) > 0. Then we can define a probability measure Q(·) = P(·|D) on F. By continuity, Q(X (τ ) = Y (τ )) = 1. For any t ⩾ 0, we can therefore write (almost surely with respect to Q) ∫ t+τ Y (t + τ ) − X (t + τ ) = Y (τ ) − X (τ ) + c(s, Y (s)) − a(s, X (s), Z (s)) ds τ ∫ t = c(s + τ, Y (s + τ )) − a(s + τ, X (s + τ ), Z (s + τ )) ds. 0

The right-hand side is continuously differentiable in t and so   Y (t + τ ) − X (t + τ ) Q lim = c(τ, X (τ )) − a(τ, X (τ ), Z (τ )) = 1. t→0 t

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Therefore, Q(Y (t + τ ) > X (t + τ ) for all sufficiently small t > 0) = 1. But due to continuity of X and Y and the definition of τ , this probability should be zero. This contradiction arises from the assumption that P(D) > 0. Therefore, P(τ = +∞) = 1.  References [1] M. Allman, Chains of interacting Brownian particles under strain, Ph.D. Thesis, Warwick University, 2010. [2] M. Allman, V. Betz, Breaking the chain, Stochastic Process. Appl. 119 (8) (2009) 2645–2659. [3] W.J. Anderson, Local behaviour of solutions of stochastic integral equations, Trans. Amer. Math. Soc. 164 (1972) 309–321. [4] G.I. Bell, Models for the specific adhesion of cells to cells, Science 200 (1978) 618–627. [5] N. Berglund, B. Gentz, Pathwise description of dynamic pitchfork bifurcations with additive noise, Probab. Theory Related Fields 122 (3) (2002) 341–388. [6] N. Berglund, B. Gentz, Noise-induced phenomena in slow–fast dynamical systems, in: Probability and its Applications (New York), Springer-Verlag London Ltd., London, 2006. [7] A. Bovier, M. Eckhoff, V. Gayrard, M. Klein, Metastability in reversible diffusion processes. I. Sharp asymptotics for capacities and exit times, J. Eur. Math. Soc. (JEMS) 6 (4) (2004) 399–424. [8] Digital library of mathematical functions, National Institute of Standards and Technology, 2010-05-07, URL http://dlmf.nist.gov/. [9] O.K. Dudko, A.E. Filippov, J. Klafter, M. Urbakh, Beyond the conventional description of dynamic force spectroscopy of adhesion bonds, PNAS 100 (20) (2003) 11378–11381. [10] E. Evans, K. Ritchie, Dynamic strength of molecular adhesion bonds, Biophys. J. 72 (4) (1997) 1541–1555. [11] H. Eyring, The activated complex in chemical reactions, J. Chem. Phys. 3 (1935) 107–115. [12] M.I. Freidlin, A.D. Wentzell, Random Perturbations of Dynamical Systems, second ed., vol. 260, Springer-Verlag, New York, 1998. [13] R.W. Friddle, Unified model of dynamic forced barrier crossing in single molecules, Phys. Rev. Lett. 100 (13) (2008) 138302. [14] S. Fugmann, I.M. Sokolov, Scaling of the rupture dynamics of polymer chains pulled at one end at a constant rate, Phys. Rev. E 79 (2) (2009) 021803. [15] S. Fugmann, I.M. Sokolov, Non-monotonic dependence of the polymer rupture force on molecule chain length, EPL 86 (28001) (2009) 1–5. [16] H.A. Kramers, Brownian motion in a field of force and the diffusion model of chemical reactions, Physica (Utrecht) 7 (1940) 284–304. [17] J.S. Langer, Statistical theory of the decay of metastable states, Ann. Phys. 54 (2) (1969) 258–275. [18] C.F. Lee, Thermal breakage of a discrete one-dimensional string, Phys. Rev. E 80 (3) (2009) 031134. [19] H.-J. Lin, H.-Y. Chen, Y.-J. Sheng, H.-K. Tsao, Bell’s expression and the generalized Garg form for forced dissociation of a biomolecular complex, Phys. Rev. Lett. 98 (8) (2007) 088304. [20] M. Raible, M. Evstigneev, F.W. Bartels, R. Eckel, M. Nguyen-Duong, R. Merkel, R. Ros, D. Anselmetti, P. Reimann, Theoretical analysis of single-molecule force spectroscopy experiments: heterogeneity of chemical bonds, Biophys. J. 90 (2006) 3851–3864. [21] A. Sain, C.L. Dias, M. Grant, Rupture of an extended object: a many-body Kramers calculation, Phys. Rev. E 74 (4) (2006) 046111.